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If n is a positive integer, is (n+1)! - (n-1)! greater than

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If n is a positive integer, is (n+1)! - (n-1)! greater than [#permalink]  12 Jun 2005, 17:21
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If n is a positive integer, is (n+1)! - (n-1)! greater than (n+1)!/(n-1)! ?
1) Square root of n is a positive integer
2) n is divisible by 2
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[#permalink]  12 Jun 2005, 18:32
Statement (1) does not give any useful information that i can see

(n+1)!-(n-1)! is greater than (n+1)!/(n-1)! for any positive number except 1 (as (n-1)! would be 0...i think!...thus making (n+1)!/(n-1)! undefined).
If n is divisible by 2...the lowest possible value of n is two...
this is sufficient to say (n+1)!-(n-1)! > (n+1)!/(n-1)!

thus, my answer is B
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[#permalink]  12 Jun 2005, 18:39
cloudz9 wrote:
Statement (1) does not give any useful information that i can see

(n+1)!-(n-1)! is greater than (n+1)!/(n-1)! for any positive number except 1 (as (n-1)! would be 0...i think!...thus making (n+1)!/(n-1)! undefined).
If n is divisible by 2...the lowest possible value of n is two...
this is sufficient to say (n+1)!-(n-1)! > (n+1)!/(n-1)!

thus, my answer is B

the whole equation reduces to (n-1)! * (n(n+1) -1) ... this needs to be compared with (n+1)!/(n-1)! = n(n+1)

I) Square root of n is a positive integer
n can be 1 or 2 or any number greater than 2

for n =1 0! =1 (not zero) - LHS will be less than RHS
for n = 2, LHS < RHS
for n - 3, LHS = 2*11 ; RHS = 12 => LHS > RHS
and for any number greater than 2, LHS > RHS

We need (2) to determine that n is greater than 2

(2) only this is not sufficient either as n can be equal to 2

and hence C
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[#permalink]  12 Jun 2005, 18:50
yeah true sorry
made a stupid mistake

agree with C then
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[#permalink]  12 Jun 2005, 21:45
yah, agree with C. if n <3, second term is larger. if n>2, the first term is larger.

togather, n >2. so it is sufficient.

sparky nice problem..............
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[#permalink]  13 Jun 2005, 07:12
How is it C! I am coming up with E

we need to know that if N is positive, greater than 2.....

I know it is positive from both statements, but dont know if it is greater than 2?
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[#permalink]  13 Jun 2005, 09:53
I will go with C too
The equation reduces to (n-1)![n(n+1) -1] > n(n+1)
From statement 1: n can be 1, 4, 9,...
when n is 1: (n-1)![n(n+1) -1] is not > than n(n+1)
when n is 4: (n-1)![n(n+1) -1] is > than n(n+1)
when n is 9: (n-1)![n(n+1) -1] is > than n(n+1)
so statement 1 is not sufficient

From statement 2: n can be any even number such as 2, 4, 6, 8....
when n is 2: (n-1)![n(n+1) -1] is not > than n(n+1)
when n is 4: (n-1)![n(n+1) -1] is > than n(n+1)
so statement 2 is not sufficient

But both put together: n can be 4, 16, .....
As shown above the condition is satisfied for any number greater than 2
Hence C
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[#permalink]  13 Jun 2005, 13:29
The two values can be equal, or one can be greater than the other.

For n=1, the values are equal

For n=2, LHS < RHS

For n > 2 LHS < RHS

So we need to find if n=1 or n=2 or n > 2

I alone gives us not enough information. n could be 1,4,9 etc.

II alone gives us not enough information n could be 2,4,6 etc.

I&II together say that n could be 4,16,36 etc. which is sufficient to determine that LHS < RHS

So I'll go for C
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[#permalink]  14 Jun 2005, 04:12
I will pick B.

Statement 1, it repets nothing but the original condition that n is a positive integer. It behaves differently for 1 and any higher positive integer. Hence Insufficient. Hence rule out option D as well.

Statement 2, says n is even. substraction operation on two positive integer ( Large positive integer- small positive integer) is always greater than division operation on of the same two numbers ( large positive number/small positive number). Hence B is sufficient.

So answer is B. No need to look at C and E.
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Re: DS X [#permalink]  14 Jun 2005, 10:52
sparky wrote:
If n is a positive integer, is (n+1)! - (n-1)! greater than (n+1)!/(n-1)! ?
1) Square root of n is a positive integer
2) n is divisible by 2

I think it is C.

The stem reduces to (n-1)! [ n^2 + n - 1 ] > [ n^2 + n ] ?

We need to know that if (n-1)! = 1 then it is NO , but if (n-1) ! > 1 then YES.

Since the least n which is a sqaure of a number and a multiple of 2 is 4. (n-1)! > 1 ,hece the aswer is YES always.

HMTG.
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[#permalink]  14 Jun 2005, 23:50
Quote:
tarungmat wrote:

substraction operation on two positive integer ( Large positive integer- small positive integer) is always greater than division operation on of the same two numbers ( large positive number/small positive number).

3 - 2 is not greater than 3 /2

Clarification: I meant in question's context (with factorial ). we will never have a situation 3-2 and 3/2 according to statement 2.
Please let me know why not B is correct as per my logic.
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Re: DS X [#permalink]  15 Jun 2005, 20:51
sparky wrote:
If n is a positive integer, is (n+1)! - (n-1)! greater than (n+1)!/(n-1)! ?
1) Square root of n is a positive integer
2) n is divisible by 2

(n+1)! - (n-1)!=(n-1)!*n*(n+1)-(n-1)!
(n+1)!/(n-1)! =n*(n+1)
(n+1)! - (n-1)! - (n+1)!/(n-1)! = (n-1)!*[n(n+1)-1]-n(n+1)

When n<=2, (n-1)!<=1
(n-1)!*[n(n+1)-1]-n(n+1)<=n(n+1)-1-n(n+1)=-1<0

When n>=3, (n-1)!>=2
(n-1)!*[n(n+1)-1]-n(n+1)>=2n(n+1)-1-n(n+1)=n(n+1)-1>0

(1) n is square of an integer, n could be 1 or 4 or greater, insufficient
(2) n is even, n could be 2 or 4 or greater, insufficient
Combined, n could only be 4 or greater, sufficient. (It's always >0 when n>=3)

(C)
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[#permalink]  15 Jun 2005, 21:00
Quote:
Square root of n is a positive integer
n can be 1 or 2 or any number greater than 2

How can N be 2..N can be 1or 4 or 9
Sq root of 2 is 1.41414 some thing which is not a positive integer

correct me if i am wrong...

Cheers
[#permalink] 15 Jun 2005, 21:00
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